Find the sum of all positive integral values of n for which \(\frac{n+6}{n}\) is an integer.

MobiusLoops Sep 13, 2018

#1**0 **

n can be any multiple of 6 so the sum is infinite.

This answer is compeletely wrong so skip it.

Melody Sep 13, 2018

#3**+8 **

**Find the sum of all positive integral values of n for which**

**\(\mathbf{\large\dfrac{n+6}{n}}\) **

**is an integer.**

**\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{n+6}{n}} \\\\ &=& \dfrac{n}{n} + \dfrac{6}{n}\\\\ &=& 1 + \dfrac{6}{n}\\ \hline \end{array}\)**

So \(n\) are the divisors of \(6\).

The 4 divisors of 6 are: \(1,2,3,6\)

The sum is \(1+2+3+6 =\mathbf{ 12}\)

heureka Sep 14, 2018

#4**+3 **

Listfor(n, 1, 10, (n+6) mod n) =(0, 0, 0, 2, 1, 0, 6, 6, 6, 6). Only 4 numbers that give "0" as the remainder satisfy the condition, as heureka found. They are: 1 + 2 + 3 + 6 =12. After 6, all other numbers will give a remainder of 6.

Guest Sep 14, 2018

#6

#7**0 **

Hi MobiusLoops,

I do not believe you are using the word correctly.

This is an **Integer** or integer value question.

Integrals are completely different. They are a part of calculus and nothing to do with integer values.

I understand the confusion, Integral is an normal English word and integral and integer would have the same origin in speech. But in maths the two meanings are very different.

Melody
Sep 14, 2018

#8**+1 **

Melody, I don't think so. My math question stated that question sooo... I don't know if they made a mistake or I'm not sure.

MobiusLoops Sep 17, 2018